DISINI 2011EuclidSolution
The CENTRE for EDUCATION in
MATHEMATICS and COMPUTING
2011 Euclid Contest
Tuesday, April 12, 2011
Solutions
©2011 Centre for Education in Mathematics and Computing
2011 Euclid Contest Solutions
Page 2
1. (a) Since (x + 1) + (x + 2) + (x + 3) = 8 + 9 + 10, then 3x + 6 = 27 or 3x = 21 and so x = 7.
p
√
√
√
(b) Since √ 25 + x = 6, then squaring both sides gives 25 + x = 36 or x = 11.
Since x =p11, then squaring both sides again, we obtain x = 112 = 121.
√
√
√
Checking, 25 + 121 = 25 + 11 = 36 = 6, as required.
(c) Since (a, 2) is the point of intersection of the lines with equations y = 2x−4 and y = x+k,
then the coordinates of this point must satisfy both equations.
Using the first equation, 2 = 2a − 4 or 2a = 6 or a = 3.
Since the coordinates of the point (3, 2) satisfy the equation y = x + k, then 2 = 3 + k or
k = −1.
2. (a) Since the side length of the original square is 3 and an equilateral triangle of side length 1
is removed from the middle of each side, then each of the two remaining pieces of each
side of the square has length 1.
Also, each of the two sides of each of the equilateral triangles that are shown has length 1.
1
1
1
1
Therefore, each of the 16 line segments in the figure has length 1, and so the perimeter of
the figure is 16.
(b) Since DC = DB, then △CDB is isosceles and ∠DBC = ∠DCB = 15◦ .
Thus, ∠CDB = 180◦ − ∠DBC − ∠DCB = 150◦ .
Since the angles around a point add to 360◦ , then
∠ADC = 360◦ − ∠ADB − ∠CDB = 360◦ − 130◦ − 150◦ = 80◦ .
(c) By the Pythagorean
Theorem in △EAD, we have EA2 +AD2 = ED2 or 122 +AD2 = 132 ,
√
and so AD = 169 − 144 = 5, since AD > 0.
By the Pythagorean
Theorem in △ACD, we have AC 2 + CD2 = AD2 or AC 2 + 42 = 52 ,
√
and so AC = 25 − 16 = 3, since AC > 0.
(We could also have determined the lengths of AD and AC by recognizing 3-4-5 and
5-12-13 right-angled triangles.)
By the Pythagorean
Theorem
in △ABC, we have AB 2 + BC 2 = AC 2 or AB 2 + 22 = 32 ,
√
√
and so AB = 9 − 4 = 5, since AB > 0.
3. (a) Solution 1
y
Since we want to make 15 − as large as possible, then we want to subtract as little as
x
possible from 15.
y
In other words, we want to make as small as possible.
x
To make a fraction with positive numerator and denominator as small as possible, we
make the numerator as small as possible and the denominator as large as possible.
Since 2 ≤ x ≤ 5 and 10 ≤ y ≤ 20, then we make x = 5 and y = 10.
y
10
Therefore, the maximum value of 15 − is 15 −
= 13.
x
5
2011 Euclid Contest Solutions
Page 3
Solution 2
y
y
Since y is positive and 2 ≤ x ≤ 5, then 15 − ≤ 15 − for any x with 2 ≤ x ≤ 5 and
x
5
positive y.
10
y
for any y with 10 ≤ y ≤ 20.
Since 10 ≤ y ≤ 20, then 15 − ≤ 15 −
5
5
y
10
Therefore, for any x and y in these ranges, 15 − ≤ 15 −
= 13, and so the maximum
x
5
possible value is 13 (which occurs when x = 5 and y = 10).
(b) Solution 1
First, we add the two given equations to obtain
(f (x) + g(x)) + (f (x) − g(x)) = (3x + 5) + (5x + 7)
or 2f (x) = 8x + 12 which gives f (x) = 4x + 6.
Since f (x) + g(x) = 3x + 5, then g(x) = 3x + 5 − f (x) = 3x + 5 − (4x + 6) = −x − 1.
(We could also find g(x) by subtracting the two given equations or by using the second of
the given equations.)
Since f (x) = 4x + 6, then f (2) = 14.
Since g(x) = −x − 1, then g(2) = −3.
Therefore, 2f (2)g(2) = 2 × 14 × (−3) = −84.
Solution 2
Since the two given equations are true for all values of x, then we can substitute x = 2 to
obtain
f (2) + g(2) = 11
f (2) − g(2) = 17
Next, we add these two equations to obtain 2f (2) = 28 or f (2) = 14.
Since f (2) + g(2) = 11, then g(2) = 11 − f (2) = 11 − 14 = −3.
(We could also find g(2) by subtracting the two equations above or by using the second
of these equations.)
Therefore, 2f (2)g(2) = 2 × 14 × (−3) = −84.
4. (a) We consider choosing the three numbers all at once.
We list the possible sets of three numbers that can be chosen:
{1, 2, 3} {1, 2, 4} {1, 2, 5} {1, 3, 4} {1, 3, 5} {1, 4, 5} {2, 3, 4} {2, 3, 5} {2, 4, 5} {3, 4, 5}
We have listed each in increasing order because once the numbers are chosen, we arrange
them in increasing order.
There are 10 sets of three numbers that can be chosen.
Of these 10, the 4 sequences 1, 2, 3 and 1, 3, 5 and 2, 3, 4 and 3, 4, 5 are arithmetic sequences.
4
or 52 .
Therefore, the probability that the resulting sequence is an arithmetic sequence is 10
2011 Euclid Contest Solutions
Page 4
(b) Solution 1
Join B to D.
B
A
D
60
C
Consider △CBD.
Since CB = CD, then ∠CBD = ∠CDB = 12 (180◦ − ∠BCD) = 21 (180◦ − 60◦ ) = 60◦ .
Therefore, △BCD is equilateral, and so BD = BC = CD = 6.
Consider △DBA.
Note that ∠DBA = 90◦ − ∠CBD = 90◦ − 60◦ = 30◦ .
Since BD = BA = 6, then ∠BDA = ∠BAD = 21 (180◦ − ∠DBA) = 12 (180◦ − 30◦ ) = 75◦ .
We calculate the length of AD.
Method 1
AD
BA
=
.
sin(∠DBA)
sin(∠BDA)
6 × 12
3
6 sin(30◦ )
=
=
.
Therefore, AD =
◦
◦
sin(75 )
sin(75 )
sin(75◦ )
By the Sine Law in △DBA, we have
Method 2
If we drop a perpendicular from B to P on AD, then P is the midpoint of AD since
△BDA is isosceles. Thus, AD = 2AP .
Also, BP bisects ∠DBA, so ∠ABP = 15◦ .
Now, AP = BA sin(∠ABP ) = 6 sin(15◦ ).
Therefore, AD = 2AP = 12 sin(15◦ ).
Method 3
By the Cosine Law in △DBA,
AD2 = AB 2 + BD2 − 2(AB)(BD) cos(∠ABD)
= 62 + 62 − 2(6)(6) cos(30◦ )
√
= 72 − 72( 23 )
√
= 72 − 36 3
q
p
√
√
Therefore, AD = 36(2 − 3) = 6 2 − 3 since AD > 0.
2011 Euclid Contest Solutions
Page 5
Solution 2
Drop perpendiculars from D to Q on BC and from D to R on BA.
R
B
Q
A
D
60
C
Then CQ = CD cos(∠DCQ) = 6 cos(60◦ ) = 6 × √21 = 3.
√
Also, DQ = CD sin(∠DCQ) = 6 sin(60◦ ) = 6 × 23 = 3 3.
Since BC = 6, then BQ = BC − CQ = 6 − 3 = 3.
Now quadrilateral BQDR has three right angles, so it must have a fourth right angle and
so must be a rectangle.
√
Thus, RD = BQ = 3 and RB = DQ = 3 3. √
Since AB = 6, then AR = AB − RB = 6 − 3 3.
Since △ARD is right-angled at R, then using the Pythagorean Theorem and the fact that
AD > 0, we obtain
q
q
q
√
√
√
√
2
2
2
2
AD = RD + AR = 3 + (6 − 3 3) = 9 + 36 − 36 3 + 27 = 72 − 36 3
q
p
√
√
which we can rewrite as AD = 36(2 − 3) = 6 2 − 3.
5. (a) Let n be the original number and N be the number when the digits are reversed. Since
we are looking for the largest value of n, we assume that n > 0.
Since we want N to be 75% larger than n, then N should be 175% of n, or N = 47 n.
Suppose that the tens digit of n is a and the units digit of n is b. Then n = 10a + b.
Also, the tens digit of N is b and the units digit of N is a, so N = 10b + a.
We want 10b + a = 47 (10a + b) or 4(10b + a) = 7(10a + b) or 40b + 4a = 70a + 7b or
33b = 66a, and so b = 2a.
This tells us that that any two-digit number n = 10a + b with b = 2a has the required
property.
Since both a and b are digits then b < 10 and so a < 5, which means that the possible
values of n are 12, 24, 36, and 48.
The largest of these numbers is 48.
(b) We “complete the rectangle” by drawing a horizontal line through C which meets the
y-axis at P and the vertical line through B at Q.
y
P
C (k, 5)
Q
A (0, 3)
O
B (4, 0)
x
2011 Euclid Contest Solutions
Page 6
Since C has y-coordinate 5, then P has y-coordinate 5; thus the coordinates of P are
(0, 5).
Since B has x-coordinate 4, then Q has x-coordinate 4.
Since C has y-coordinate 5, then Q has y-coordinate 5.
Therefore, the coordinates of Q are (4, 5), and so rectangle OP QB is 4 by 5 and so has
area 4 × 5 = 20.
Now rectangle OP QB is made up of four smaller triangles, and so the sum of the areas of
these triangles must be 20.
Let us examine each of these triangles:
• △ABC has area 8 (given information)
• △AOB is right-angled at O, has height AO = 3 and base OB = 4, and so has area
1
× 4 × 3 = 6.
2
• △AP C is right-angled at P , has height AP = 5 − 3 = 2 and base P C = k − 0 = k,
and so has area 12 × k × 2 = k.
• △CQB is right-angled at Q, has height QB = 5 − 0 = 5 and base CQ = 4 − k, and
so has area 21 × (4 − k) × 5 = 10 − 25 k.
Since the sum of the areas of these triangles is 20, then 8 + 6 + k + 10 − 52 k = 20 or 4 = 23 k
and so k = 83 .
6. (a) Solution 1
Suppose that the distance from point A to point B is d km.
Suppose also that rc is the speed at which Serge travels while not paddling (i.e. being
carried by just the current), that rp is the speed at which Serge travels with no current
(i.e. just from his paddling), and rp+c his speed when being moved by both his paddling
and the current.
It takes Serge 18 minutes to travel from A to B while paddling with the current.
d
km/min.
Thus, rp+c =
18
It takes Serge 30 minutes to travel from A to B with just the current.
d
Thus, rc =
km/min.
30
d
5d 3d
2d
d
d
−
=
−
=
=
km/min.
But rp = rp+c − rc =
18 30
90 90
90
45
d
Since Serge can paddle the d km from A to B at a speed of
km/min, then it takes him
45
45 minutes to paddle from A to B with no current.
Solution 2
Suppose that the distance from point A to point B is d km, the speed of the current of
the river is r km/h, and the speed that Serge can paddle is s km/h.
d
1
1
Since the current can carry Serge from A to B in 30 minutes (or h), then = .
2
r
2
When Serge paddles with the current, his speed equals his paddling speed plus the speed
of the current, or (s + r) km/h.
3
Since Serge can paddle with the current from A to B in 18 minutes (or
h), then
10
d
3
= .
r+s
10
d
The time to paddle from A to B with no current would be h.
s
2011 Euclid Contest Solutions
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1
r
d
= , then = 2.
r
2
d
3
r+s
10
d
= , then
= .
Since
r+s
10
d
3
s
r+s r
10
4
− =
−2= .
Therefore, =
d
d
d
3
3
d
3
3
Thus, = , and so it would take Serge of an hour, or 45 minutes, to paddle from A
s
4
4
to B with no current.
Since
Solution 3
Suppose that the distance from point A to point B is d km, the speed of the current of
the river is r km/h, and the speed that Serge can paddle is s km/h.
d
1
1
= or
Since the current can carry Serge from A to B in 30 minutes (or h), then
2
r
2
d = 21 r.
When Serge paddles with the current, his speed equals his paddling speed plus the speed
of the current, or (s + r) km/h.
3
Since Serge can paddle with the current from A to B in 18 minutes (or
h), then
10
3
d
3
=
or d = 10
(r + s).
r+s
10
3
3
Since d = 21 r and d = 10
(r + s), then 21 r = 10
(r + s) or 5r = 3r + 3s and so s = 23 r.
1
r
3
d
2
To travel from A to B with no current, the time in hours that it takes is = 2 = , or
s
4
r
3
45 minutes.
(b) First, we note that a 6= 0. (If a = 0, then the “parabola” y = a(x − 2)(x − 6) is actually
the horizontal line y = 0 which intersects the square all along OR.)
Second, we note that, regardless of the value of a 6= 0, the parabola has x-intercepts 2 and
6, and so intersects the x-axis at (2, 0) and (6, 0), which we call K(2, 0) and L(6, 0). This
gives KL = 4.
Third, we note that since the x-intercepts of the parabola are 2 and 6, then the axis of
symmetry of the parabola has equation x = 21 (2 + 6) = 4.
Since the axis of symmetry of the parabola is a vertical line of symmetry, then if the
parabola intersects the two vertical sides of the square, it will intersect these at the same
height, and if the parabola intersects the top side of the square, it will intersect it at two
points that are symmetrical about the vertical line x = 4.
Fourth, we recall that a trapezoid with parallel sides of lengths a and b and height h has
area 12 h(a + b).
We now examine three cases.
2011 Euclid Contest Solutions
Page 8
Case 1: a < 0
Here, the parabola opens downwards.
Since the parabola intersects the square at four points, it must intersect P Q at points
M and N . (The parabola cannot intersect the vertical sides of the square since it gets
“narrower” towards the vertex.)
x=4
y
N
P
M
K
Q (8, 8)
L
R
O
x
Since the parabola opens downwards, then M N < KL = 4.
Since the height of the trapezoid equals the height of the square (or 8), then the area of
the trapezoid is 12 h(KL + M N ) which is less than 21 (8)(4 + 4) = 32.
But the area of the trapezoid must be 36, so this case is not possible.
Case 2: a > 0; M and N on P Q
We have the following configuration:
x=4
y
P
N
K
O
Q (8, 8)
M
L
R
x
Here, the height of the trapezoid is 8, KL = 4, and M and N are symmetric about x = 4.
Since the area of the trapezoid is 36, then 21 h(KL + M N ) = 36 or 21 (8)(4 + M N ) = 36 or
4 + M N = 9 or M N = 5.
Thus, M and N are each 25 units from x = 4, and so N has coordinates ( 32 , 8).
Since this point lies on the parabola with equation y = a(x − 2)(x − 6), then
.
8 = a( 32 − 2)( 32 − 6) or 8 = a(− 12 )(− 29 ) or 8 = 94 a or a = 32
9
2011 Euclid Contest Solutions
Page 9
Case 3: a > 0; M and N on QR and P O
We have the following configuration:
x=4
y
Q
P
N
O
M
L
K
R
x
Here, KL = 4, M N = 8, and M and N have the same y-coordinate.
Since the area of the trapezoid is 36, then 21 h(KL + M N ) = 36 or 12 h(4 + 8) = 36 or
6h = 36 or h = 6.
Thus, N has coordinates (0, 6).
Since this point lies on the parabola with equation y = a(x − 2)(x − 6), then
6 = a(0 − 2)(0 − 6) or 6 = 12a or a = 21 .
Therefore, the possible values of a are
32
9
and 12 .
7. (a) Solution 1
Consider a population of 100 people, each of whom is 75 years old and who behave according to the probabilities given in the question.
Each of the original 100 people has a 50% chance of living at least another 10 years, so
there will be 50% × 100 = 50 of these people alive at age 85.
Each of the original 100 people has a 20% chance of living at least another 15 years, so
there will be 20% × 100 = 20 of these people alive at age 90.
Since there is a 25% (or 41 ) chance that an 80 year old person will live at least another 10
years (that is, to age 90), then there should be 4 times as many of these people alive at
age 80 than at age 90.
Since there are 20 people alive at age 90, then there are 4 × 20 = 80 of the original 100
people alive at age 80.
In summary, of the initial 100 people of age 75, there are 80 alive at age 80, 50 alive at
age 85, and 20 people alive at age 90.
Because 50 of the 80 people alive at age 80 are still alive at age 85, then the probability
50
that an 80 year old person will live at least 5 more years (that is, to age 85) is 80
= 58 , or
62.5%.
Solution 2
Suppose that the probability that a 75 year old person lives to 80 is p, the probability
that an 80 year old person lives to 85 is q, and the probability that an 85 year old person
lives to 90 is r.
We want to the determine the value of q.
For a 75 year old person to live at least another 10 years, they must live another 5 years
(to age 80) and then another 5 years (to age 85). The probability of this is equal to pq.
We are told in the question that this is equal to 50% or 0.5.
Therefore, pq = 0.5.
2011 Euclid Contest Solutions
Page 10
For a 75 year old person to live at least another 15 years, they must live another 5 years
(to age 80), then another 5 years (to age 85), and then another 5 years (to age 90). The
probability of this is equal to pqr. We are told in the question that this is equal to 20%
or 0.2.
Therefore, pqr = 0.2
Similarly, since the probability that an 80 year old person will live another 10 years is
25%, then qr = 0.25.
0.2
pqr
=
= 0.4.
Since pqr = 0.2 and pq = 0.5, then r =
pq
0.5
0.25
qr
Since qr = 0.25 and r = 0.4, then q =
=
= 0.625.
r
0.4
Therefore, the probability that an 80 year old man will live at least another 5 years is
0.625, or 62.5%.
(b) Using logarithm rules, the given equation is equivalent to 22 log10 x = 3(2 · 2log10 x ) + 16 or
(2log10 x )2 = 6 · 2log10 x + 16.
Set u = 2log10 x . Then the equation becomes u2 = 6u + 16 or u2 − 6u − 16 = 0.
Factoring, we obtain (u − 8)(u + 2) = 0 and so u = 8 or u = −2.
Since 2a > 0 for any real number a, then u > 0 and so we can reject the possibility that
u = −2.
Thus, u = 2log10 x = 8 which means that log10 x = 3.
Therefore, x = 1000.
8. (a) First, we determine the first entry in the 50th row.
Since the first column is an arithmetic sequence with common difference 3, then the 50th
entry in the first column (the first entry in the 50th row) is 4 + 49(3) = 4 + 147 = 151.
Second, we determine the common difference in the 50th row by determining the second
entry in the 50th row.
Since the second column is an arithmetic sequence with common difference 5, then the
50th entry in the second column (that is, the second entry in the 50th row) is 7 + 49(5)
or 7 + 245 = 252.
Therefore, the common difference in the 50th row must be 252 − 151 = 101.
Thus, the 40th entry in the 50th row (that is, the number in the 50th row and the 40th
column) is 151 + 39(101) = 151 + 3939 = 4090.
(b) We follow the same procedure as in (a).
First, we determine the first entry in the Rth row.
Since the first column is an arithmetic sequence with common difference 3, then the Rth
entry in the first column (that is, the first entry in the Rth row) is 4 + (R − 1)(3) or
4 + 3R − 3 = 3R + 1.
Second, we determine the common difference in the Rth row by determining the second
entry in the Rth row.
Since the second column is an arithmetic sequence with common difference 5, then the
Rth entry in the second column (that is, the second entry in the Rth row) is 7 + (R − 1)(5)
or 7 + 5R − 5 = 5R + 2.
Therefore, the common difference in the Rth row must be (5R + 2) − (3R + 1) = 2R + 1.
Thus, the Cth entry in the Rth row (that is, the number in the Rth row and the Cth
column) is
3R + 1 + (C − 1)(2R + 1) = 3R + 1 + 2RC + C − 2R − 1 = 2RC + R + C
2011 Euclid Contest Solutions
Page 11
(c) Suppose that N is an entry in the table, say in the Rth row and Cth column.
From (b), then N = 2RC + R + C and so 2N + 1 = 4RC + 2R + 2C + 1.
Now 4RC + 2R + 2C + 1 = 2R(2C + 1) + 2C + 1 = (2R + 1)(2C + 1).
Since R and C are integers with R ≥ 1 and C ≥ 1, then 2R + 1 and 2C + 1 are each
integers that are at least 3.
Therefore, 2N + 1 = (2R + 1)(2C + 1) must be composite, since it is the product of two
integers that are each greater than 1.
√
9. (a) If n = 2011,
√ then 8n − 7 = 16081 and so 8n − 7 ≈ 126.81.
1 + 8n − 7
1 + 126.81
Thus,
≈
≈ 63.9.
2
2 $
%
p
1 + 8(2011) − 7
= 4022 + ⌊63.9⌋ = 4022 + 63 = 4085.
Therefore, g(2011) = 2(2011) +
2
(b) To determine a value of n for which f (n) = 100, we need to solve the equation
√
1 + 8n − 7
2n −
= 100
(∗)
2
We first solve the equation
2x −
1+
√
8x − 7
= 100
2
(∗∗)
because the left sides of (∗) and (∗∗) do not differ by much and so the solutions are likely
close together. We will try integers n in (∗) that are close to the solutions to (∗∗).
Manipulating (∗∗), we obtain
√
4x − (1 + 8x − 7) = 200
√
4x − 201 =
8x − 7
2
(4x − 201) = 8x − 7
16x2 − 1608x + 40401 = 8x − 7
16x2 − 1616x + 40408 = 0
2x2 − 202x + 5051 = 0
By the quadratic formula,
p
√
√
202 ± 2022 − 4(2)(5051)
202 ± 396
101 ± 99
=
=
x=
2(2)
4
2
and so x ≈ 55.47 or x ≈ 45.53.
We try n = 55, which is close to 55.47:
$
%
$
%
p
√
1 + 8(55) − 7
1 + 433
f (55) = 2(55) −
= 110 −
2
2
√
√
433
≈ 10.9, which gives
2
Thus, f (55) = 110 − 10 = 100.
Therefore, a value of n for which f (n) = 100 is n = 55.
Since
433 ≈ 20.8, then
1+
$
1+
√
2
433
%
= 10.
2011 Euclid Contest Solutions
Page 12
(c) We want to show that each positive integer m is in the range of f or the range of g, but
not both.
To do this, we first try to better understand the “complicated” term of each of the functions – that is, the term involving the greatest integer function.
In particular, we start
integer k ≥ 1 and try to determine the positive
√ a positive
with
1 + 8n − 7
= k.
integers n that give
2
√
1 + 8n − 7
= k is equivBy definition of the greatest integer function, the equation
2
√
1 + 8n − 7
alent to the inequality k ≤
< k + 1, from which we obtain the following set
2
of equivalent inequalities
√
2k + 2
2k
≤ 1+
√ 8n − 7 <
2k − 1
≤
8n − 7
<
2k + 1
2
2
4k − 4k + 1 ≤
8n − 7
< 4k + 4k + 1
4k 2 − 4k + 8 ≤
8n
< 4k 2 + 4k + 8
1
(k 2 − k) + 1 ≤
n
< 12 (k 2 + k) + 1
2
If we define Tk = 21 k(k + 1) = 12 (k 2 + k) to be the kth triangular number for k ≥ 0, then
Tk−1 = 21 (k − 1)(k) = 12 (k 2 − k).
√
1 + 8n − 7
= k for Tk−1 + 1 ≤ n < Tk + 1.
Therefore,
2
√
1 + 8n − 7
= k is true for Tk−1 + 1 ≤ n ≤ Tk .
Since n is an integer, then
2
When k = 1, this interval is T0 + 1 ≤ n ≤ T1 (or 1 ≤ n ≤ 1). When k = 2, this interval
is T1 + 1 ≤ n ≤ T2 (or 2 ≤ n ≤ 3). When k = 3, this interval is T2 + 1 ≤ n ≤ T3 (or
4 ≤ n ≤ 6). As k ranges over all positive integers, these intervals include every positive
integer n and do not overlap.
Therefore, we can determine the range of each of the functions f and g by examining the
values f (n) and g(n) when n is in these intervals.
For each non-negative integer k, define Rk to be the set of integers greater than k 2 and
less than or equal to (k + 1)2 . Thus, Rk = {k 2 + 1, k 2 + 2, . . . , k 2 + 2k, k 2 + 2k + 1}.
For example, R0 = {1}, R1 = {2, 3, 4}, R2 = {5, 6, 7, 8, 9}, and so on. Every positive
integer occurs in exactly one of these sets.
Also, for each non-negative integer k define Sk = {k 2 + 2, k 2 + 4, . . . , k 2 + 2k} and define
Qk = {k 2 + 1, k 2 + 3, . . . , k 2 + 2k + 1}. For example, S0 = {}, S1 = {3}, S2 = {6, 8},
Q0 = {1}, Q1 = {2, 4}, Q2 = {5, 7, 9}, and so on. Note that Rk = Qk ∪ Sk so every
positive integer occurs in exactly one Qk or in exactly one Sk , and that these sets do not
overlap since no two Sk ’s overlap and no two Qk ’s overlap and no Qk overlaps with an Sk .
We determine the range of the function g first.
√
1 + 8n − 7
For Tk−1 + 1 ≤ n ≤ Tk , we have
= k and so
2
2Tk−1 + 2
≤
2n√
1 + 8n − 7
2Tk−1 + 2 + k ≤ 2n +
2
k2 − k + 2 + k ≤
g(n)
k2 + 2
≤
g(n)
≤
2Tk
≤
2Tk + k
≤ k2 + k + k
≤ k 2 + 2k
2011 Euclid Contest Solutions
Page 13
Note that when n is in this interval and increases by 1, then the 2n term causes the value
of g(n) to increase by 2.
Therefore, for the values of n in this interval, g(n) takes precisely the values k 2 + 2,
k 2 + 4, k 2 + 6, . . . , k 2 + 2k.
In other words, the range of g over this interval of its domain is precisely the set Sk .
As k ranges over all positive integers (that is, as these intervals cover the domain of g),
this tells us that the range of g is precisely the integers in the sets S1 , S2 , S3 , . . ..
(We could also include S0 in this list since it is the empty set.)
√
1+ 8−7
We note next that f (1) = 2 −
= 1, the only element of Q0 .
2
√
1 + 8n − 7
= k + 1 and so
For k ≥ 1 and Tk + 1 ≤ n ≤ Tk+1 , we have
2
2Tk + 2
≤
2n√
1 + 8n − 7
2Tk + 2 − (k + 1) ≤ 2n −
2
k2 + k + 2 − k − 1 ≤
f (n)
k2 + 1
≤
f (n)
≤
2Tk+1
≤
2Tk+1 − (k + 1)
≤ (k + 1)(k + 2) − k − 1
≤
k 2 + 2k + 1
Note that when n is in this interval and increases by 1, then the 2n term causes the value
of f (n) to increase by 2.
Therefore, for the values of n in this interval, f (n) takes precisely the values k 2 + 1,
k 2 + 3, k 2 + 5, . . . , k 2 + 2k + 1.
In other words, the range of f over this interval of its domain is precisely the set Qk .
As k ranges over all positive integers (that is, as these intervals cover the domain of f ),
this tells us that the range of f is precisely the integers in the sets Q0 , Q1 , Q2 , . . ..
Therefore, the range of f is the set of elements in the sets Q0 , Q1 , Q2 , . . . and the range
of g is the set of elements in the sets S0 , S1 , S2 , . . .. These ranges include every positive
integer and do not overlap.
10. (a) Suppose that ∠KAB = θ.
Since ∠KAC = 2∠KAB, then ∠KAC = 2θ and ∠BAC = ∠KAC + ∠KAB = 3θ.
Since 3∠ABC = 2∠BAC, then ∠ABC = 23 × 3θ = 2θ.
Since ∠AKC is exterior to △AKB, then ∠AKC = ∠KAB + ∠ABC = 3θ.
This gives the following configuration:
C
a
x
b
3θ
2θ
A
K
d
x
2θ
θ
c
B
Now △CAK is similar to △CBA since the triangles have a common angle at C and
∠CAK = ∠CBA.
2011 Euclid Contest Solutions
Page 14
CA
d
b
bc
AK
=
or = and so d = .
BA
CB
c
a
a
CK
CA
a−x
b
b2
b2
a2 − b 2
Also,
=
or
= and so a − x =
or x = a −
=
, as required.
CA
CB
b
a
a
a
a
(b) From (a), bc = ad and a2 − b2 = ax and so we obtain
Therefore,
LS = (a2 − b2 )(a2 − b2 + ac) = (ax)(ax + ac) = a2 x(x + c)
and
RS = b2 c2 = (bc)2 = (ad)2 = a2 d2
In order to show that LS = RS, we need to show that x(x + c) = d2 (since a > 0).
Method 1: Use the Sine Law
First, we derive a formula for sin 3θ which we will need in this solution:
sin 3θ =
=
=
=
=
sin(2θ + θ)
sin 2θ cos θ + cos 2θ sin θ
2 sin θ cos2 θ + (1 − 2 sin2 θ) sin θ
2 sin θ(1 − sin2 θ) + (1 − 2 sin2 θ) sin θ
3 sin θ − 4 sin3 θ
Since ∠AKB = 180◦ − ∠KAB − ∠KBA = 180◦ − 3θ, then using the Sine Law in △AKB
gives
x
d
c
=
=
sin θ
sin 2θ
sin(180◦ − 3θ)
d sin θ
and
Since sin(180◦ − X) = sin X, then sin(180◦ − 3θ) = sin 3θ, and so x =
sin 2θ
d sin 3θ
. This gives
c=
sin 2θ
d sin θ d sin θ d sin 3θ
x(x + c) =
+
sin 2θ sin 2θ
sin 2θ
2
d sin θ
(sin θ + sin 3θ)
=
sin2 2θ
d2 sin θ
=
(sin θ + 3 sin θ − 4 sin3 θ)
sin2 2θ
d2 sin θ
=
(4 sin θ − 4 sin3 θ)
2
sin 2θ
4d2 sin2 θ
=
(1 − sin2 θ)
sin2 2θ
4d2 sin2 θ cos2 θ
=
sin2 2θ
2
4d sin2 θ cos2 θ
=
(2 sin θ cos θ)2
4d2 sin2 θ cos2 θ
=
4 sin2 θ cos2 θ
2
= d
2011 Euclid Contest Solutions
Page 15
as required.
We could have instead used the formula sin A + sin B = 2 sin
show that sin 3θ + sin θ = 2 sin 2θ cos θ, from which
A+B
2
cos
A−B
2
to
sin θ(sin 3θ + sin θ) = sin θ(2 sin 2θ cos θ) = 2 sin θ cos θ sin 2θ = sin2 2θ
Method 2: Extend AB
Extend AB to E so that BE = BK = x and join KE.
C
a x
b
3θ
2θ
A
K
d
x
2θ
θ
c
x
B
E
Now △KBE is isosceles with ∠BKE = ∠KEB.
Since ∠KBA is the exterior angle of △KBE, then ∠KBA = 2∠KEB = 2θ.
Thus, ∠KEB = ∠BKE = θ.
But this also tells us that ∠KAE = ∠KEA = θ.
Thus, △KAE is isosceles and so KE = KA = d.
C
a
x
b
3θ
2θ
A
d
K
d
x
2θ
θ
c
θ
B
x
E
So △KAE is similar to △BKE, since each has two angles equal to θ.
KA
AE
d
c+x
Thus,
=
or =
and so d2 = x(x + c), as required.
BK
KE
x
d
Method 3: Use the Cosine Law and the Sine Law
We apply the Cosine Law in △AKB to obtain
AK 2 = BK 2 + BA2 − 2(BA)(BK) cos(∠KBA)
d2 = x2 + c2 − 2cx cos(2θ)
d2 = x2 + c2 − 2cx(2 cos2 θ − 1)
d
sin 2θ
d
2 sin θ cos θ
d
x
=
or
= or
=
Using the Sine Law in △AKB, we get
sin θ
sin 2θ
sin θ
x
sin θ
x
d
.
and so cos θ =
2x
2011 Euclid Contest Solutions
Page 16
Combining these two equations,
d
2
2
2
= x + c − 2cx
d 2 = x2 + c2 −
cd2
x
2
cd
d2 +
x
2
xd + cd2
d2 (x + c)
d2
d2 +
2d2
−1
4x2
cd2
+ 2cx
x
= x2 + 2cx + c2
= (x + c)2
= x(x + c)2
= x(x + c)2
= x(x + c)
as required (since x + c 6= 0).
(c) Solution 1
Our goal is to find a triple of positive integers that satisfy the equation in (b) and are the
side lengths of a triangle.
First, we note that if (A, B, C) is a triple of real numbers that satisfies the equation in
(b) and k is another real number, then the triple (kA, kB, kC) also satisfies the equation
from (b), since
(k 2 A2 −k 2 B 2 )(k 2 A2 −k 2 B 2 +kAkC) = k 4 (A2 −B 2 )(A2 −B 2 +AC) = k 4 (B 2 C 2 ) = (kB)2 (kC)2
Therefore, we start by trying to find a triple (a, b, c) of rational numbers that satisfies
the equation in (b) and forms a triangle, and then “scale up” this triple to form a triple
(ka, kb, kc) of integers.
To do this, we rewrite the equation from (b) as a quadratic equation in c and solve for c
using the quadratic formula.
Partially expanding the left side from (b), we obtain
(a2 − b2 )(a2 − b2 ) + ac(a2 − b2 ) = b2 c2
which we rearrange to obtain
b2 c2 − c(a(a2 − b2 )) − (a2 − b2 )2 = 0
By the quadratic formula,
p
p
a(a2 − b2 ) ± a2 (a2 − b2 )2 + 4b2 (a2 − b2 )2
a(a2 − b2 ) ± (a2 − b2 )2 (a2 + 4b2 )
c=
=
2b2
2b2
Since ∠BAC > ∠ABC, then a > b and so a2 − b2 > 0, which gives
√
√
(a2 − b2 )
a(a2 − b2 ) ± (a2 − b2 ) a2 + 4b2
a2 + 4b2 )
=
(a
±
c=
2b2
2b2
√
Since a2 + 4b2 > 0, then a2 + 4b2 > a, so the positive root is
c=
p
(a2 − b2 )
(a
+
a2 + (2b)2 )
2b2
2011 Euclid Contest Solutions
Page 17
We try to find integers a and b that give a rational value for c. We will then check to see
if this triple (a, b, c) forms the side lengths of a triangle, and then eventually scale these
up to get integer values.
p
One way for the value of c to be rational (and in fact the only way) is for a2 + (2b)2 to
be an √
integer, or for a and 2b to be the legs of a Pythagorean triple.
Since 32 + 42 is an integer, then we try a = 3 and b = 2, which gives
c=
√
(32 − 22 )
(3
+
32 + 4 2 ) = 5
2 · 22
and so (a, b, c) = (3, 2, 5). Unfortunately, these lengths do not form a triangle, since
3 + 2 = 5.
(The Triangle Inequality tells us that three positive real numbers a, b and c form a triangle
if and only if a + b > c and a + c > b and b + c > a.)
We can continue to try small Pythagorean triples.
Now 152 + 82 = 172 , but a = 15 and b = 4 do not give a value of c that forms a triangle
with a and b.
However, 162 + 302 = 342 , so we can try a = 16 and b = 15 which gives
c=
√
31
(162 − 152 )
31
(16 + 162 + 302 ) =
(16 + 34) =
2
2 · 15
450
9
) do form the sides of a triangle since a + b > c and
Now the lengths (a, b, c) = (16, 15, 31
9
a + c > b and b + c > a.
Since these values satisfy the equation from (b), then we can scale them up by a factor of
k = 9 to obtain the triple (144, 135, 31) which satisfies the equation from (b) and are the
side lengths of a triangle.
(Using other Pythagorean triples, we could obtain other triples of integers that work.)
Solution 2
We note that the equation in (b) involves only a, b and c and so appears to depend only
on the relationship between the angles ∠CAB and ∠CBA in △ABC.
Using this premise, we use △ABC, remove the line segment AK and draw the altitude
CF .
C
a
b
2θ
3θ
A
b cos 3θ
F
a cos 2θ
B
Because we are only looking for one triple that works, we can make a number of assumptions that may or may not be true in general for such a triangle, but which will help us
find an example.
We assume that 3θ and 2θ are both acute angles; that is, we assume that θ < 30◦ .
In △ABC, we have AF = b cos 3θ, BF = a cos 2θ, and CF = b sin 3θ = a sin 2θ.
Note also that c = b cos 3θ + a cos 2θ.
2011 Euclid Contest Solutions
Page 18
One way to find the integers a, b, c that we require is to look for integers a and b and an
angle θ with the properties that b cos 3θ and a cos 2θ are integers and b sin 3θ = a sin 2θ.
Using trigonometric formulae,
sin 2θ = 2 sin θ cos θ
cos 2θ = 2 cos2 θ − 1
sin 3θ = 3 sin θ − 4 sin3 θ
(from the calculation in (a), Solution 1, Method 1)
cos 3θ = cos(2θ + θ)
= cos 2θ cos θ − sin 2θ sin θ
= (2 cos2 θ − 1) cos θ − 2 sin2 θ cos θ
= (2 cos2 θ − 1) cos θ − 2(1 − cos2 θ) cos θ
= 4 cos3 θ − 3 cos θ
So we can try to find an angle θ < 30◦ with cos θ a rational number and then integers a
and b that make b sin 3θ = a sin 2θ and ensure that√b cos 3θ and a cos 2θ are integers.
Since we are assuming that θ < 30◦ , then cos θ > 23 ≈ 0.866.
√
The rational number with smallest denominator that is larger than 23 is 78 , so we try the
acute angle θ with cos θ = 87 .
√
√
In this case, sin θ = 1 − cos2 θ = 815 , and so
√
√
sin 2θ = 2 sin θ cos θ = 2 × 87 × 815 = 7 3215
17
− 1 = 32
cos 2θ = 2 cos2 θ − 1 = 2 × 49
64
sin 3θ = 3 sin θ − 4 sin3 θ = 3 ×
cos 3θ = 4 cos3 θ − 3 cos θ = 4 ×
√
√
√
15
8
343
512
−4×
−3×
√
√
15 15
33 15
=
512
128
7
7
= 128
8
To have b sin 3θ = a sin 2θ, we need 3312815 b = 7 3215 a or 33b = 28a.
17
7
b and 32
a to be integers, and
To ensure that b cos 3θ and a cos 2θ are integers, we need 128
so a must be divisible by 32 and b must be divisible by 128.
The integers a = 33 and b = 28 satisfy the equation 33b = 28a.
Multiplying each by 32 gives a = 1056 and b = 896 which satisfy the equation 33b = 28a
and now have the property that b is divisible by 128 (with quotient 7) and a is divisible
by 32 (with quotient 33).
7
17
With these values of a and b, we obtain c = b cos 3θ+a cos 2θ = 896× 128
+1056× 32
= 610.
We can then check that the triple (a, b, c) = (1056, 896, 610) satisfies the equation from
(b), as required.
As in our discussion in Solution 1, each element of this triple can be divided by 2 to obtain
the “smaller” triple (a, b, c) = (528, 448, 305) that satisfies the equation too.
Using other values for cos θ and integers a and b, we could obtain other triples (a, b, c) of
integers that work.
MATHEMATICS and COMPUTING
2011 Euclid Contest
Tuesday, April 12, 2011
Solutions
©2011 Centre for Education in Mathematics and Computing
2011 Euclid Contest Solutions
Page 2
1. (a) Since (x + 1) + (x + 2) + (x + 3) = 8 + 9 + 10, then 3x + 6 = 27 or 3x = 21 and so x = 7.
p
√
√
√
(b) Since √ 25 + x = 6, then squaring both sides gives 25 + x = 36 or x = 11.
Since x =p11, then squaring both sides again, we obtain x = 112 = 121.
√
√
√
Checking, 25 + 121 = 25 + 11 = 36 = 6, as required.
(c) Since (a, 2) is the point of intersection of the lines with equations y = 2x−4 and y = x+k,
then the coordinates of this point must satisfy both equations.
Using the first equation, 2 = 2a − 4 or 2a = 6 or a = 3.
Since the coordinates of the point (3, 2) satisfy the equation y = x + k, then 2 = 3 + k or
k = −1.
2. (a) Since the side length of the original square is 3 and an equilateral triangle of side length 1
is removed from the middle of each side, then each of the two remaining pieces of each
side of the square has length 1.
Also, each of the two sides of each of the equilateral triangles that are shown has length 1.
1
1
1
1
Therefore, each of the 16 line segments in the figure has length 1, and so the perimeter of
the figure is 16.
(b) Since DC = DB, then △CDB is isosceles and ∠DBC = ∠DCB = 15◦ .
Thus, ∠CDB = 180◦ − ∠DBC − ∠DCB = 150◦ .
Since the angles around a point add to 360◦ , then
∠ADC = 360◦ − ∠ADB − ∠CDB = 360◦ − 130◦ − 150◦ = 80◦ .
(c) By the Pythagorean
Theorem in △EAD, we have EA2 +AD2 = ED2 or 122 +AD2 = 132 ,
√
and so AD = 169 − 144 = 5, since AD > 0.
By the Pythagorean
Theorem in △ACD, we have AC 2 + CD2 = AD2 or AC 2 + 42 = 52 ,
√
and so AC = 25 − 16 = 3, since AC > 0.
(We could also have determined the lengths of AD and AC by recognizing 3-4-5 and
5-12-13 right-angled triangles.)
By the Pythagorean
Theorem
in △ABC, we have AB 2 + BC 2 = AC 2 or AB 2 + 22 = 32 ,
√
√
and so AB = 9 − 4 = 5, since AB > 0.
3. (a) Solution 1
y
Since we want to make 15 − as large as possible, then we want to subtract as little as
x
possible from 15.
y
In other words, we want to make as small as possible.
x
To make a fraction with positive numerator and denominator as small as possible, we
make the numerator as small as possible and the denominator as large as possible.
Since 2 ≤ x ≤ 5 and 10 ≤ y ≤ 20, then we make x = 5 and y = 10.
y
10
Therefore, the maximum value of 15 − is 15 −
= 13.
x
5
2011 Euclid Contest Solutions
Page 3
Solution 2
y
y
Since y is positive and 2 ≤ x ≤ 5, then 15 − ≤ 15 − for any x with 2 ≤ x ≤ 5 and
x
5
positive y.
10
y
for any y with 10 ≤ y ≤ 20.
Since 10 ≤ y ≤ 20, then 15 − ≤ 15 −
5
5
y
10
Therefore, for any x and y in these ranges, 15 − ≤ 15 −
= 13, and so the maximum
x
5
possible value is 13 (which occurs when x = 5 and y = 10).
(b) Solution 1
First, we add the two given equations to obtain
(f (x) + g(x)) + (f (x) − g(x)) = (3x + 5) + (5x + 7)
or 2f (x) = 8x + 12 which gives f (x) = 4x + 6.
Since f (x) + g(x) = 3x + 5, then g(x) = 3x + 5 − f (x) = 3x + 5 − (4x + 6) = −x − 1.
(We could also find g(x) by subtracting the two given equations or by using the second of
the given equations.)
Since f (x) = 4x + 6, then f (2) = 14.
Since g(x) = −x − 1, then g(2) = −3.
Therefore, 2f (2)g(2) = 2 × 14 × (−3) = −84.
Solution 2
Since the two given equations are true for all values of x, then we can substitute x = 2 to
obtain
f (2) + g(2) = 11
f (2) − g(2) = 17
Next, we add these two equations to obtain 2f (2) = 28 or f (2) = 14.
Since f (2) + g(2) = 11, then g(2) = 11 − f (2) = 11 − 14 = −3.
(We could also find g(2) by subtracting the two equations above or by using the second
of these equations.)
Therefore, 2f (2)g(2) = 2 × 14 × (−3) = −84.
4. (a) We consider choosing the three numbers all at once.
We list the possible sets of three numbers that can be chosen:
{1, 2, 3} {1, 2, 4} {1, 2, 5} {1, 3, 4} {1, 3, 5} {1, 4, 5} {2, 3, 4} {2, 3, 5} {2, 4, 5} {3, 4, 5}
We have listed each in increasing order because once the numbers are chosen, we arrange
them in increasing order.
There are 10 sets of three numbers that can be chosen.
Of these 10, the 4 sequences 1, 2, 3 and 1, 3, 5 and 2, 3, 4 and 3, 4, 5 are arithmetic sequences.
4
or 52 .
Therefore, the probability that the resulting sequence is an arithmetic sequence is 10
2011 Euclid Contest Solutions
Page 4
(b) Solution 1
Join B to D.
B
A
D
60
C
Consider △CBD.
Since CB = CD, then ∠CBD = ∠CDB = 12 (180◦ − ∠BCD) = 21 (180◦ − 60◦ ) = 60◦ .
Therefore, △BCD is equilateral, and so BD = BC = CD = 6.
Consider △DBA.
Note that ∠DBA = 90◦ − ∠CBD = 90◦ − 60◦ = 30◦ .
Since BD = BA = 6, then ∠BDA = ∠BAD = 21 (180◦ − ∠DBA) = 12 (180◦ − 30◦ ) = 75◦ .
We calculate the length of AD.
Method 1
AD
BA
=
.
sin(∠DBA)
sin(∠BDA)
6 × 12
3
6 sin(30◦ )
=
=
.
Therefore, AD =
◦
◦
sin(75 )
sin(75 )
sin(75◦ )
By the Sine Law in △DBA, we have
Method 2
If we drop a perpendicular from B to P on AD, then P is the midpoint of AD since
△BDA is isosceles. Thus, AD = 2AP .
Also, BP bisects ∠DBA, so ∠ABP = 15◦ .
Now, AP = BA sin(∠ABP ) = 6 sin(15◦ ).
Therefore, AD = 2AP = 12 sin(15◦ ).
Method 3
By the Cosine Law in △DBA,
AD2 = AB 2 + BD2 − 2(AB)(BD) cos(∠ABD)
= 62 + 62 − 2(6)(6) cos(30◦ )
√
= 72 − 72( 23 )
√
= 72 − 36 3
q
p
√
√
Therefore, AD = 36(2 − 3) = 6 2 − 3 since AD > 0.
2011 Euclid Contest Solutions
Page 5
Solution 2
Drop perpendiculars from D to Q on BC and from D to R on BA.
R
B
Q
A
D
60
C
Then CQ = CD cos(∠DCQ) = 6 cos(60◦ ) = 6 × √21 = 3.
√
Also, DQ = CD sin(∠DCQ) = 6 sin(60◦ ) = 6 × 23 = 3 3.
Since BC = 6, then BQ = BC − CQ = 6 − 3 = 3.
Now quadrilateral BQDR has three right angles, so it must have a fourth right angle and
so must be a rectangle.
√
Thus, RD = BQ = 3 and RB = DQ = 3 3. √
Since AB = 6, then AR = AB − RB = 6 − 3 3.
Since △ARD is right-angled at R, then using the Pythagorean Theorem and the fact that
AD > 0, we obtain
q
q
q
√
√
√
√
2
2
2
2
AD = RD + AR = 3 + (6 − 3 3) = 9 + 36 − 36 3 + 27 = 72 − 36 3
q
p
√
√
which we can rewrite as AD = 36(2 − 3) = 6 2 − 3.
5. (a) Let n be the original number and N be the number when the digits are reversed. Since
we are looking for the largest value of n, we assume that n > 0.
Since we want N to be 75% larger than n, then N should be 175% of n, or N = 47 n.
Suppose that the tens digit of n is a and the units digit of n is b. Then n = 10a + b.
Also, the tens digit of N is b and the units digit of N is a, so N = 10b + a.
We want 10b + a = 47 (10a + b) or 4(10b + a) = 7(10a + b) or 40b + 4a = 70a + 7b or
33b = 66a, and so b = 2a.
This tells us that that any two-digit number n = 10a + b with b = 2a has the required
property.
Since both a and b are digits then b < 10 and so a < 5, which means that the possible
values of n are 12, 24, 36, and 48.
The largest of these numbers is 48.
(b) We “complete the rectangle” by drawing a horizontal line through C which meets the
y-axis at P and the vertical line through B at Q.
y
P
C (k, 5)
Q
A (0, 3)
O
B (4, 0)
x
2011 Euclid Contest Solutions
Page 6
Since C has y-coordinate 5, then P has y-coordinate 5; thus the coordinates of P are
(0, 5).
Since B has x-coordinate 4, then Q has x-coordinate 4.
Since C has y-coordinate 5, then Q has y-coordinate 5.
Therefore, the coordinates of Q are (4, 5), and so rectangle OP QB is 4 by 5 and so has
area 4 × 5 = 20.
Now rectangle OP QB is made up of four smaller triangles, and so the sum of the areas of
these triangles must be 20.
Let us examine each of these triangles:
• △ABC has area 8 (given information)
• △AOB is right-angled at O, has height AO = 3 and base OB = 4, and so has area
1
× 4 × 3 = 6.
2
• △AP C is right-angled at P , has height AP = 5 − 3 = 2 and base P C = k − 0 = k,
and so has area 12 × k × 2 = k.
• △CQB is right-angled at Q, has height QB = 5 − 0 = 5 and base CQ = 4 − k, and
so has area 21 × (4 − k) × 5 = 10 − 25 k.
Since the sum of the areas of these triangles is 20, then 8 + 6 + k + 10 − 52 k = 20 or 4 = 23 k
and so k = 83 .
6. (a) Solution 1
Suppose that the distance from point A to point B is d km.
Suppose also that rc is the speed at which Serge travels while not paddling (i.e. being
carried by just the current), that rp is the speed at which Serge travels with no current
(i.e. just from his paddling), and rp+c his speed when being moved by both his paddling
and the current.
It takes Serge 18 minutes to travel from A to B while paddling with the current.
d
km/min.
Thus, rp+c =
18
It takes Serge 30 minutes to travel from A to B with just the current.
d
Thus, rc =
km/min.
30
d
5d 3d
2d
d
d
−
=
−
=
=
km/min.
But rp = rp+c − rc =
18 30
90 90
90
45
d
Since Serge can paddle the d km from A to B at a speed of
km/min, then it takes him
45
45 minutes to paddle from A to B with no current.
Solution 2
Suppose that the distance from point A to point B is d km, the speed of the current of
the river is r km/h, and the speed that Serge can paddle is s km/h.
d
1
1
Since the current can carry Serge from A to B in 30 minutes (or h), then = .
2
r
2
When Serge paddles with the current, his speed equals his paddling speed plus the speed
of the current, or (s + r) km/h.
3
Since Serge can paddle with the current from A to B in 18 minutes (or
h), then
10
d
3
= .
r+s
10
d
The time to paddle from A to B with no current would be h.
s
2011 Euclid Contest Solutions
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1
r
d
= , then = 2.
r
2
d
3
r+s
10
d
= , then
= .
Since
r+s
10
d
3
s
r+s r
10
4
− =
−2= .
Therefore, =
d
d
d
3
3
d
3
3
Thus, = , and so it would take Serge of an hour, or 45 minutes, to paddle from A
s
4
4
to B with no current.
Since
Solution 3
Suppose that the distance from point A to point B is d km, the speed of the current of
the river is r km/h, and the speed that Serge can paddle is s km/h.
d
1
1
= or
Since the current can carry Serge from A to B in 30 minutes (or h), then
2
r
2
d = 21 r.
When Serge paddles with the current, his speed equals his paddling speed plus the speed
of the current, or (s + r) km/h.
3
Since Serge can paddle with the current from A to B in 18 minutes (or
h), then
10
3
d
3
=
or d = 10
(r + s).
r+s
10
3
3
Since d = 21 r and d = 10
(r + s), then 21 r = 10
(r + s) or 5r = 3r + 3s and so s = 23 r.
1
r
3
d
2
To travel from A to B with no current, the time in hours that it takes is = 2 = , or
s
4
r
3
45 minutes.
(b) First, we note that a 6= 0. (If a = 0, then the “parabola” y = a(x − 2)(x − 6) is actually
the horizontal line y = 0 which intersects the square all along OR.)
Second, we note that, regardless of the value of a 6= 0, the parabola has x-intercepts 2 and
6, and so intersects the x-axis at (2, 0) and (6, 0), which we call K(2, 0) and L(6, 0). This
gives KL = 4.
Third, we note that since the x-intercepts of the parabola are 2 and 6, then the axis of
symmetry of the parabola has equation x = 21 (2 + 6) = 4.
Since the axis of symmetry of the parabola is a vertical line of symmetry, then if the
parabola intersects the two vertical sides of the square, it will intersect these at the same
height, and if the parabola intersects the top side of the square, it will intersect it at two
points that are symmetrical about the vertical line x = 4.
Fourth, we recall that a trapezoid with parallel sides of lengths a and b and height h has
area 12 h(a + b).
We now examine three cases.
2011 Euclid Contest Solutions
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Case 1: a < 0
Here, the parabola opens downwards.
Since the parabola intersects the square at four points, it must intersect P Q at points
M and N . (The parabola cannot intersect the vertical sides of the square since it gets
“narrower” towards the vertex.)
x=4
y
N
P
M
K
Q (8, 8)
L
R
O
x
Since the parabola opens downwards, then M N < KL = 4.
Since the height of the trapezoid equals the height of the square (or 8), then the area of
the trapezoid is 12 h(KL + M N ) which is less than 21 (8)(4 + 4) = 32.
But the area of the trapezoid must be 36, so this case is not possible.
Case 2: a > 0; M and N on P Q
We have the following configuration:
x=4
y
P
N
K
O
Q (8, 8)
M
L
R
x
Here, the height of the trapezoid is 8, KL = 4, and M and N are symmetric about x = 4.
Since the area of the trapezoid is 36, then 21 h(KL + M N ) = 36 or 21 (8)(4 + M N ) = 36 or
4 + M N = 9 or M N = 5.
Thus, M and N are each 25 units from x = 4, and so N has coordinates ( 32 , 8).
Since this point lies on the parabola with equation y = a(x − 2)(x − 6), then
.
8 = a( 32 − 2)( 32 − 6) or 8 = a(− 12 )(− 29 ) or 8 = 94 a or a = 32
9
2011 Euclid Contest Solutions
Page 9
Case 3: a > 0; M and N on QR and P O
We have the following configuration:
x=4
y
Q
P
N
O
M
L
K
R
x
Here, KL = 4, M N = 8, and M and N have the same y-coordinate.
Since the area of the trapezoid is 36, then 21 h(KL + M N ) = 36 or 12 h(4 + 8) = 36 or
6h = 36 or h = 6.
Thus, N has coordinates (0, 6).
Since this point lies on the parabola with equation y = a(x − 2)(x − 6), then
6 = a(0 − 2)(0 − 6) or 6 = 12a or a = 21 .
Therefore, the possible values of a are
32
9
and 12 .
7. (a) Solution 1
Consider a population of 100 people, each of whom is 75 years old and who behave according to the probabilities given in the question.
Each of the original 100 people has a 50% chance of living at least another 10 years, so
there will be 50% × 100 = 50 of these people alive at age 85.
Each of the original 100 people has a 20% chance of living at least another 15 years, so
there will be 20% × 100 = 20 of these people alive at age 90.
Since there is a 25% (or 41 ) chance that an 80 year old person will live at least another 10
years (that is, to age 90), then there should be 4 times as many of these people alive at
age 80 than at age 90.
Since there are 20 people alive at age 90, then there are 4 × 20 = 80 of the original 100
people alive at age 80.
In summary, of the initial 100 people of age 75, there are 80 alive at age 80, 50 alive at
age 85, and 20 people alive at age 90.
Because 50 of the 80 people alive at age 80 are still alive at age 85, then the probability
50
that an 80 year old person will live at least 5 more years (that is, to age 85) is 80
= 58 , or
62.5%.
Solution 2
Suppose that the probability that a 75 year old person lives to 80 is p, the probability
that an 80 year old person lives to 85 is q, and the probability that an 85 year old person
lives to 90 is r.
We want to the determine the value of q.
For a 75 year old person to live at least another 10 years, they must live another 5 years
(to age 80) and then another 5 years (to age 85). The probability of this is equal to pq.
We are told in the question that this is equal to 50% or 0.5.
Therefore, pq = 0.5.
2011 Euclid Contest Solutions
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For a 75 year old person to live at least another 15 years, they must live another 5 years
(to age 80), then another 5 years (to age 85), and then another 5 years (to age 90). The
probability of this is equal to pqr. We are told in the question that this is equal to 20%
or 0.2.
Therefore, pqr = 0.2
Similarly, since the probability that an 80 year old person will live another 10 years is
25%, then qr = 0.25.
0.2
pqr
=
= 0.4.
Since pqr = 0.2 and pq = 0.5, then r =
pq
0.5
0.25
qr
Since qr = 0.25 and r = 0.4, then q =
=
= 0.625.
r
0.4
Therefore, the probability that an 80 year old man will live at least another 5 years is
0.625, or 62.5%.
(b) Using logarithm rules, the given equation is equivalent to 22 log10 x = 3(2 · 2log10 x ) + 16 or
(2log10 x )2 = 6 · 2log10 x + 16.
Set u = 2log10 x . Then the equation becomes u2 = 6u + 16 or u2 − 6u − 16 = 0.
Factoring, we obtain (u − 8)(u + 2) = 0 and so u = 8 or u = −2.
Since 2a > 0 for any real number a, then u > 0 and so we can reject the possibility that
u = −2.
Thus, u = 2log10 x = 8 which means that log10 x = 3.
Therefore, x = 1000.
8. (a) First, we determine the first entry in the 50th row.
Since the first column is an arithmetic sequence with common difference 3, then the 50th
entry in the first column (the first entry in the 50th row) is 4 + 49(3) = 4 + 147 = 151.
Second, we determine the common difference in the 50th row by determining the second
entry in the 50th row.
Since the second column is an arithmetic sequence with common difference 5, then the
50th entry in the second column (that is, the second entry in the 50th row) is 7 + 49(5)
or 7 + 245 = 252.
Therefore, the common difference in the 50th row must be 252 − 151 = 101.
Thus, the 40th entry in the 50th row (that is, the number in the 50th row and the 40th
column) is 151 + 39(101) = 151 + 3939 = 4090.
(b) We follow the same procedure as in (a).
First, we determine the first entry in the Rth row.
Since the first column is an arithmetic sequence with common difference 3, then the Rth
entry in the first column (that is, the first entry in the Rth row) is 4 + (R − 1)(3) or
4 + 3R − 3 = 3R + 1.
Second, we determine the common difference in the Rth row by determining the second
entry in the Rth row.
Since the second column is an arithmetic sequence with common difference 5, then the
Rth entry in the second column (that is, the second entry in the Rth row) is 7 + (R − 1)(5)
or 7 + 5R − 5 = 5R + 2.
Therefore, the common difference in the Rth row must be (5R + 2) − (3R + 1) = 2R + 1.
Thus, the Cth entry in the Rth row (that is, the number in the Rth row and the Cth
column) is
3R + 1 + (C − 1)(2R + 1) = 3R + 1 + 2RC + C − 2R − 1 = 2RC + R + C
2011 Euclid Contest Solutions
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(c) Suppose that N is an entry in the table, say in the Rth row and Cth column.
From (b), then N = 2RC + R + C and so 2N + 1 = 4RC + 2R + 2C + 1.
Now 4RC + 2R + 2C + 1 = 2R(2C + 1) + 2C + 1 = (2R + 1)(2C + 1).
Since R and C are integers with R ≥ 1 and C ≥ 1, then 2R + 1 and 2C + 1 are each
integers that are at least 3.
Therefore, 2N + 1 = (2R + 1)(2C + 1) must be composite, since it is the product of two
integers that are each greater than 1.
√
9. (a) If n = 2011,
√ then 8n − 7 = 16081 and so 8n − 7 ≈ 126.81.
1 + 8n − 7
1 + 126.81
Thus,
≈
≈ 63.9.
2
2 $
%
p
1 + 8(2011) − 7
= 4022 + ⌊63.9⌋ = 4022 + 63 = 4085.
Therefore, g(2011) = 2(2011) +
2
(b) To determine a value of n for which f (n) = 100, we need to solve the equation
√
1 + 8n − 7
2n −
= 100
(∗)
2
We first solve the equation
2x −
1+
√
8x − 7
= 100
2
(∗∗)
because the left sides of (∗) and (∗∗) do not differ by much and so the solutions are likely
close together. We will try integers n in (∗) that are close to the solutions to (∗∗).
Manipulating (∗∗), we obtain
√
4x − (1 + 8x − 7) = 200
√
4x − 201 =
8x − 7
2
(4x − 201) = 8x − 7
16x2 − 1608x + 40401 = 8x − 7
16x2 − 1616x + 40408 = 0
2x2 − 202x + 5051 = 0
By the quadratic formula,
p
√
√
202 ± 2022 − 4(2)(5051)
202 ± 396
101 ± 99
=
=
x=
2(2)
4
2
and so x ≈ 55.47 or x ≈ 45.53.
We try n = 55, which is close to 55.47:
$
%
$
%
p
√
1 + 8(55) − 7
1 + 433
f (55) = 2(55) −
= 110 −
2
2
√
√
433
≈ 10.9, which gives
2
Thus, f (55) = 110 − 10 = 100.
Therefore, a value of n for which f (n) = 100 is n = 55.
Since
433 ≈ 20.8, then
1+
$
1+
√
2
433
%
= 10.
2011 Euclid Contest Solutions
Page 12
(c) We want to show that each positive integer m is in the range of f or the range of g, but
not both.
To do this, we first try to better understand the “complicated” term of each of the functions – that is, the term involving the greatest integer function.
In particular, we start
integer k ≥ 1 and try to determine the positive
√ a positive
with
1 + 8n − 7
= k.
integers n that give
2
√
1 + 8n − 7
= k is equivBy definition of the greatest integer function, the equation
2
√
1 + 8n − 7
alent to the inequality k ≤
< k + 1, from which we obtain the following set
2
of equivalent inequalities
√
2k + 2
2k
≤ 1+
√ 8n − 7 <
2k − 1
≤
8n − 7
<
2k + 1
2
2
4k − 4k + 1 ≤
8n − 7
< 4k + 4k + 1
4k 2 − 4k + 8 ≤
8n
< 4k 2 + 4k + 8
1
(k 2 − k) + 1 ≤
n
< 12 (k 2 + k) + 1
2
If we define Tk = 21 k(k + 1) = 12 (k 2 + k) to be the kth triangular number for k ≥ 0, then
Tk−1 = 21 (k − 1)(k) = 12 (k 2 − k).
√
1 + 8n − 7
= k for Tk−1 + 1 ≤ n < Tk + 1.
Therefore,
2
√
1 + 8n − 7
= k is true for Tk−1 + 1 ≤ n ≤ Tk .
Since n is an integer, then
2
When k = 1, this interval is T0 + 1 ≤ n ≤ T1 (or 1 ≤ n ≤ 1). When k = 2, this interval
is T1 + 1 ≤ n ≤ T2 (or 2 ≤ n ≤ 3). When k = 3, this interval is T2 + 1 ≤ n ≤ T3 (or
4 ≤ n ≤ 6). As k ranges over all positive integers, these intervals include every positive
integer n and do not overlap.
Therefore, we can determine the range of each of the functions f and g by examining the
values f (n) and g(n) when n is in these intervals.
For each non-negative integer k, define Rk to be the set of integers greater than k 2 and
less than or equal to (k + 1)2 . Thus, Rk = {k 2 + 1, k 2 + 2, . . . , k 2 + 2k, k 2 + 2k + 1}.
For example, R0 = {1}, R1 = {2, 3, 4}, R2 = {5, 6, 7, 8, 9}, and so on. Every positive
integer occurs in exactly one of these sets.
Also, for each non-negative integer k define Sk = {k 2 + 2, k 2 + 4, . . . , k 2 + 2k} and define
Qk = {k 2 + 1, k 2 + 3, . . . , k 2 + 2k + 1}. For example, S0 = {}, S1 = {3}, S2 = {6, 8},
Q0 = {1}, Q1 = {2, 4}, Q2 = {5, 7, 9}, and so on. Note that Rk = Qk ∪ Sk so every
positive integer occurs in exactly one Qk or in exactly one Sk , and that these sets do not
overlap since no two Sk ’s overlap and no two Qk ’s overlap and no Qk overlaps with an Sk .
We determine the range of the function g first.
√
1 + 8n − 7
For Tk−1 + 1 ≤ n ≤ Tk , we have
= k and so
2
2Tk−1 + 2
≤
2n√
1 + 8n − 7
2Tk−1 + 2 + k ≤ 2n +
2
k2 − k + 2 + k ≤
g(n)
k2 + 2
≤
g(n)
≤
2Tk
≤
2Tk + k
≤ k2 + k + k
≤ k 2 + 2k
2011 Euclid Contest Solutions
Page 13
Note that when n is in this interval and increases by 1, then the 2n term causes the value
of g(n) to increase by 2.
Therefore, for the values of n in this interval, g(n) takes precisely the values k 2 + 2,
k 2 + 4, k 2 + 6, . . . , k 2 + 2k.
In other words, the range of g over this interval of its domain is precisely the set Sk .
As k ranges over all positive integers (that is, as these intervals cover the domain of g),
this tells us that the range of g is precisely the integers in the sets S1 , S2 , S3 , . . ..
(We could also include S0 in this list since it is the empty set.)
√
1+ 8−7
We note next that f (1) = 2 −
= 1, the only element of Q0 .
2
√
1 + 8n − 7
= k + 1 and so
For k ≥ 1 and Tk + 1 ≤ n ≤ Tk+1 , we have
2
2Tk + 2
≤
2n√
1 + 8n − 7
2Tk + 2 − (k + 1) ≤ 2n −
2
k2 + k + 2 − k − 1 ≤
f (n)
k2 + 1
≤
f (n)
≤
2Tk+1
≤
2Tk+1 − (k + 1)
≤ (k + 1)(k + 2) − k − 1
≤
k 2 + 2k + 1
Note that when n is in this interval and increases by 1, then the 2n term causes the value
of f (n) to increase by 2.
Therefore, for the values of n in this interval, f (n) takes precisely the values k 2 + 1,
k 2 + 3, k 2 + 5, . . . , k 2 + 2k + 1.
In other words, the range of f over this interval of its domain is precisely the set Qk .
As k ranges over all positive integers (that is, as these intervals cover the domain of f ),
this tells us that the range of f is precisely the integers in the sets Q0 , Q1 , Q2 , . . ..
Therefore, the range of f is the set of elements in the sets Q0 , Q1 , Q2 , . . . and the range
of g is the set of elements in the sets S0 , S1 , S2 , . . .. These ranges include every positive
integer and do not overlap.
10. (a) Suppose that ∠KAB = θ.
Since ∠KAC = 2∠KAB, then ∠KAC = 2θ and ∠BAC = ∠KAC + ∠KAB = 3θ.
Since 3∠ABC = 2∠BAC, then ∠ABC = 23 × 3θ = 2θ.
Since ∠AKC is exterior to △AKB, then ∠AKC = ∠KAB + ∠ABC = 3θ.
This gives the following configuration:
C
a
x
b
3θ
2θ
A
K
d
x
2θ
θ
c
B
Now △CAK is similar to △CBA since the triangles have a common angle at C and
∠CAK = ∠CBA.
2011 Euclid Contest Solutions
Page 14
CA
d
b
bc
AK
=
or = and so d = .
BA
CB
c
a
a
CK
CA
a−x
b
b2
b2
a2 − b 2
Also,
=
or
= and so a − x =
or x = a −
=
, as required.
CA
CB
b
a
a
a
a
(b) From (a), bc = ad and a2 − b2 = ax and so we obtain
Therefore,
LS = (a2 − b2 )(a2 − b2 + ac) = (ax)(ax + ac) = a2 x(x + c)
and
RS = b2 c2 = (bc)2 = (ad)2 = a2 d2
In order to show that LS = RS, we need to show that x(x + c) = d2 (since a > 0).
Method 1: Use the Sine Law
First, we derive a formula for sin 3θ which we will need in this solution:
sin 3θ =
=
=
=
=
sin(2θ + θ)
sin 2θ cos θ + cos 2θ sin θ
2 sin θ cos2 θ + (1 − 2 sin2 θ) sin θ
2 sin θ(1 − sin2 θ) + (1 − 2 sin2 θ) sin θ
3 sin θ − 4 sin3 θ
Since ∠AKB = 180◦ − ∠KAB − ∠KBA = 180◦ − 3θ, then using the Sine Law in △AKB
gives
x
d
c
=
=
sin θ
sin 2θ
sin(180◦ − 3θ)
d sin θ
and
Since sin(180◦ − X) = sin X, then sin(180◦ − 3θ) = sin 3θ, and so x =
sin 2θ
d sin 3θ
. This gives
c=
sin 2θ
d sin θ d sin θ d sin 3θ
x(x + c) =
+
sin 2θ sin 2θ
sin 2θ
2
d sin θ
(sin θ + sin 3θ)
=
sin2 2θ
d2 sin θ
=
(sin θ + 3 sin θ − 4 sin3 θ)
sin2 2θ
d2 sin θ
=
(4 sin θ − 4 sin3 θ)
2
sin 2θ
4d2 sin2 θ
=
(1 − sin2 θ)
sin2 2θ
4d2 sin2 θ cos2 θ
=
sin2 2θ
2
4d sin2 θ cos2 θ
=
(2 sin θ cos θ)2
4d2 sin2 θ cos2 θ
=
4 sin2 θ cos2 θ
2
= d
2011 Euclid Contest Solutions
Page 15
as required.
We could have instead used the formula sin A + sin B = 2 sin
show that sin 3θ + sin θ = 2 sin 2θ cos θ, from which
A+B
2
cos
A−B
2
to
sin θ(sin 3θ + sin θ) = sin θ(2 sin 2θ cos θ) = 2 sin θ cos θ sin 2θ = sin2 2θ
Method 2: Extend AB
Extend AB to E so that BE = BK = x and join KE.
C
a x
b
3θ
2θ
A
K
d
x
2θ
θ
c
x
B
E
Now △KBE is isosceles with ∠BKE = ∠KEB.
Since ∠KBA is the exterior angle of △KBE, then ∠KBA = 2∠KEB = 2θ.
Thus, ∠KEB = ∠BKE = θ.
But this also tells us that ∠KAE = ∠KEA = θ.
Thus, △KAE is isosceles and so KE = KA = d.
C
a
x
b
3θ
2θ
A
d
K
d
x
2θ
θ
c
θ
B
x
E
So △KAE is similar to △BKE, since each has two angles equal to θ.
KA
AE
d
c+x
Thus,
=
or =
and so d2 = x(x + c), as required.
BK
KE
x
d
Method 3: Use the Cosine Law and the Sine Law
We apply the Cosine Law in △AKB to obtain
AK 2 = BK 2 + BA2 − 2(BA)(BK) cos(∠KBA)
d2 = x2 + c2 − 2cx cos(2θ)
d2 = x2 + c2 − 2cx(2 cos2 θ − 1)
d
sin 2θ
d
2 sin θ cos θ
d
x
=
or
= or
=
Using the Sine Law in △AKB, we get
sin θ
sin 2θ
sin θ
x
sin θ
x
d
.
and so cos θ =
2x
2011 Euclid Contest Solutions
Page 16
Combining these two equations,
d
2
2
2
= x + c − 2cx
d 2 = x2 + c2 −
cd2
x
2
cd
d2 +
x
2
xd + cd2
d2 (x + c)
d2
d2 +
2d2
−1
4x2
cd2
+ 2cx
x
= x2 + 2cx + c2
= (x + c)2
= x(x + c)2
= x(x + c)2
= x(x + c)
as required (since x + c 6= 0).
(c) Solution 1
Our goal is to find a triple of positive integers that satisfy the equation in (b) and are the
side lengths of a triangle.
First, we note that if (A, B, C) is a triple of real numbers that satisfies the equation in
(b) and k is another real number, then the triple (kA, kB, kC) also satisfies the equation
from (b), since
(k 2 A2 −k 2 B 2 )(k 2 A2 −k 2 B 2 +kAkC) = k 4 (A2 −B 2 )(A2 −B 2 +AC) = k 4 (B 2 C 2 ) = (kB)2 (kC)2
Therefore, we start by trying to find a triple (a, b, c) of rational numbers that satisfies
the equation in (b) and forms a triangle, and then “scale up” this triple to form a triple
(ka, kb, kc) of integers.
To do this, we rewrite the equation from (b) as a quadratic equation in c and solve for c
using the quadratic formula.
Partially expanding the left side from (b), we obtain
(a2 − b2 )(a2 − b2 ) + ac(a2 − b2 ) = b2 c2
which we rearrange to obtain
b2 c2 − c(a(a2 − b2 )) − (a2 − b2 )2 = 0
By the quadratic formula,
p
p
a(a2 − b2 ) ± a2 (a2 − b2 )2 + 4b2 (a2 − b2 )2
a(a2 − b2 ) ± (a2 − b2 )2 (a2 + 4b2 )
c=
=
2b2
2b2
Since ∠BAC > ∠ABC, then a > b and so a2 − b2 > 0, which gives
√
√
(a2 − b2 )
a(a2 − b2 ) ± (a2 − b2 ) a2 + 4b2
a2 + 4b2 )
=
(a
±
c=
2b2
2b2
√
Since a2 + 4b2 > 0, then a2 + 4b2 > a, so the positive root is
c=
p
(a2 − b2 )
(a
+
a2 + (2b)2 )
2b2
2011 Euclid Contest Solutions
Page 17
We try to find integers a and b that give a rational value for c. We will then check to see
if this triple (a, b, c) forms the side lengths of a triangle, and then eventually scale these
up to get integer values.
p
One way for the value of c to be rational (and in fact the only way) is for a2 + (2b)2 to
be an √
integer, or for a and 2b to be the legs of a Pythagorean triple.
Since 32 + 42 is an integer, then we try a = 3 and b = 2, which gives
c=
√
(32 − 22 )
(3
+
32 + 4 2 ) = 5
2 · 22
and so (a, b, c) = (3, 2, 5). Unfortunately, these lengths do not form a triangle, since
3 + 2 = 5.
(The Triangle Inequality tells us that three positive real numbers a, b and c form a triangle
if and only if a + b > c and a + c > b and b + c > a.)
We can continue to try small Pythagorean triples.
Now 152 + 82 = 172 , but a = 15 and b = 4 do not give a value of c that forms a triangle
with a and b.
However, 162 + 302 = 342 , so we can try a = 16 and b = 15 which gives
c=
√
31
(162 − 152 )
31
(16 + 162 + 302 ) =
(16 + 34) =
2
2 · 15
450
9
) do form the sides of a triangle since a + b > c and
Now the lengths (a, b, c) = (16, 15, 31
9
a + c > b and b + c > a.
Since these values satisfy the equation from (b), then we can scale them up by a factor of
k = 9 to obtain the triple (144, 135, 31) which satisfies the equation from (b) and are the
side lengths of a triangle.
(Using other Pythagorean triples, we could obtain other triples of integers that work.)
Solution 2
We note that the equation in (b) involves only a, b and c and so appears to depend only
on the relationship between the angles ∠CAB and ∠CBA in △ABC.
Using this premise, we use △ABC, remove the line segment AK and draw the altitude
CF .
C
a
b
2θ
3θ
A
b cos 3θ
F
a cos 2θ
B
Because we are only looking for one triple that works, we can make a number of assumptions that may or may not be true in general for such a triangle, but which will help us
find an example.
We assume that 3θ and 2θ are both acute angles; that is, we assume that θ < 30◦ .
In △ABC, we have AF = b cos 3θ, BF = a cos 2θ, and CF = b sin 3θ = a sin 2θ.
Note also that c = b cos 3θ + a cos 2θ.
2011 Euclid Contest Solutions
Page 18
One way to find the integers a, b, c that we require is to look for integers a and b and an
angle θ with the properties that b cos 3θ and a cos 2θ are integers and b sin 3θ = a sin 2θ.
Using trigonometric formulae,
sin 2θ = 2 sin θ cos θ
cos 2θ = 2 cos2 θ − 1
sin 3θ = 3 sin θ − 4 sin3 θ
(from the calculation in (a), Solution 1, Method 1)
cos 3θ = cos(2θ + θ)
= cos 2θ cos θ − sin 2θ sin θ
= (2 cos2 θ − 1) cos θ − 2 sin2 θ cos θ
= (2 cos2 θ − 1) cos θ − 2(1 − cos2 θ) cos θ
= 4 cos3 θ − 3 cos θ
So we can try to find an angle θ < 30◦ with cos θ a rational number and then integers a
and b that make b sin 3θ = a sin 2θ and ensure that√b cos 3θ and a cos 2θ are integers.
Since we are assuming that θ < 30◦ , then cos θ > 23 ≈ 0.866.
√
The rational number with smallest denominator that is larger than 23 is 78 , so we try the
acute angle θ with cos θ = 87 .
√
√
In this case, sin θ = 1 − cos2 θ = 815 , and so
√
√
sin 2θ = 2 sin θ cos θ = 2 × 87 × 815 = 7 3215
17
− 1 = 32
cos 2θ = 2 cos2 θ − 1 = 2 × 49
64
sin 3θ = 3 sin θ − 4 sin3 θ = 3 ×
cos 3θ = 4 cos3 θ − 3 cos θ = 4 ×
√
√
√
15
8
343
512
−4×
−3×
√
√
15 15
33 15
=
512
128
7
7
= 128
8
To have b sin 3θ = a sin 2θ, we need 3312815 b = 7 3215 a or 33b = 28a.
17
7
b and 32
a to be integers, and
To ensure that b cos 3θ and a cos 2θ are integers, we need 128
so a must be divisible by 32 and b must be divisible by 128.
The integers a = 33 and b = 28 satisfy the equation 33b = 28a.
Multiplying each by 32 gives a = 1056 and b = 896 which satisfy the equation 33b = 28a
and now have the property that b is divisible by 128 (with quotient 7) and a is divisible
by 32 (with quotient 33).
7
17
With these values of a and b, we obtain c = b cos 3θ+a cos 2θ = 896× 128
+1056× 32
= 610.
We can then check that the triple (a, b, c) = (1056, 896, 610) satisfies the equation from
(b), as required.
As in our discussion in Solution 1, each element of this triple can be divided by 2 to obtain
the “smaller” triple (a, b, c) = (528, 448, 305) that satisfies the equation too.
Using other values for cos θ and integers a and b, we could obtain other triples (a, b, c) of
integers that work.